论文标题

无嫉妒的商品,琐事和混合物品的放松

Envy-free Relaxations for Goods, Chores, and Mixed Items

论文作者

Bérczi, Kristóf, Bérczi-Kovács, Erika R., Boros, Endre, Gedefa, Fekadu Tolessa, Kamiyama, Naoyuki, Kavitha, Telikepalli, Kobayashi, Yusuke, Makino, Kazuhisa

论文摘要

在公平的部门问题中,我们将获得$ M $ $项目的$ s $ s $,以及具有个人偏好的$ n $ n $代理商的$ n $ n $ n $ n $,目标是在代理之间找到物品的分配,以便每个代理商都能找到分配公平的分配。有几个既定的公平概念,嫉妒是最广泛的研究之一。但是,当物品不可分割时,并不总是存在嫉妒的分配,这激发了嫉妒的放松:嫉妒的柔和性,最多是一项(ef1)和嫉妒的任何物品(EFX)是两个精心陈述的放松。我们考虑为不一定单调的效用函数找到EF1和EFX分配的问题,并提出了与此环境不同强度的四个可能的扩展。 特别是,我们提出了一种多项式时间算法,用于为具有任意效用函数的两种代理找到EF1分配。给出了一个示例,表明具有非符号酮,非添加,相同效用函数的两种代理不必存在EFX分配。但是,当所有代理具有单调(不一定是加性)相同的实用程序功能时,我们证明了杂务的EFX分配始终存在。为了理解一般情况,我们讨论了实用程序功能的两个子类:布尔实用程序为$ \ {0,+1 \} $ - 有价值的函数,以及$ \ {0,-1 \} $的负价函数。对于后者,我们给出了一个多项式时间算法,该算法在实用程序函数相同时找到EFX分配。

In fair division problems, we are given a set $S$ of $m$ items and a set $N$ of $n$ agents with individual preferences, and the goal is to find an allocation of items among agents so that each agent finds the allocation fair. There are several established fairness concepts and envy-freeness is one of the most extensively studied ones. However envy-free allocations do not always exist when items are indivisible and this has motivated relaxations of envy-freeness: envy-freeness up to one item (EF1) and envy-freeness up to any item (EFX) are two well-studied relaxations. We consider the problem of finding EF1 and EFX allocations for utility functions that are not necessarily monotone, and propose four possible extensions of different strength to this setting. In particular, we present a polynomial-time algorithm for finding an EF1 allocation for two agents with arbitrary utility functions. An example is given showing that EFX allocations need not exist for two agents with non-monotone, non-additive, identical utility functions. However, when all agents have monotone (not necessarily additive) identical utility functions, we prove that an EFX allocation of chores always exists. As a step toward understanding the general case, we discuss two subclasses of utility functions: Boolean utilities that are $\{0,+1\}$-valued functions, and negative Boolean utilities that are $\{0,-1\}$-valued functions. For the latter, we give a polynomial time algorithm that finds an EFX allocation when the utility functions are identical.

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